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Question 1209598: What is the domain of the function f(x) = sqrt(6x - 3 + x^2)?
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! The domain of a square root function is restricted to non-negative values under the radical. Therefore, we need to find the values of x for which:
6x - 3 + x² ≥ 0
Rearranging the terms, we get:
x² + 6x - 3 ≥ 0
To find the values of x that satisfy this inequality, we first find the roots of the corresponding quadratic equation:
x² + 6x - 3 = 0
Using the quadratic formula:
x = (-b ± √(b² - 4ac)) / 2a
x = (-6 ± √(6² - 4 * 1 * -3)) / 2 * 1
x = (-6 ± √(36 + 12)) / 2
x = (-6 ± √48) / 2
x = (-6 ± 4√3) / 2
x = -3 ± 2√3
So, the roots are x = -3 - 2√3 and x = -3 + 2√3.
Since the quadratic has a positive leading coefficient, the parabola opens upwards. This means the inequality x² + 6x - 3 ≥ 0 is satisfied when x is less than or equal to the smaller root, or x is greater than or equal to the larger root.
Therefore, the domain of f(x) is:
(-∞, -3 - 2√3] ∪ [-3 + 2√3, ∞)
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